Q45 / 150 · MHT CET 2025 (5 May Shift 2) · Chemistry

For a certain reaction $\Delta H=-225 \mathrm{~kJ}$ and $\Delta S=-150 \mathrm{JK}^{-1}$. Find the temperature so that $\Delta \mathrm{G}$ is zero.

  • Correct answer
    A
    1500 K
  • B
    1450 K
  • C
    1340 K
  • D
    1300 K

Solution

The temperature where $\Delta G$ becomes zero corresponds to the condition $\Delta G = \Delta H - T\Delta S = 0$.

For $\Delta H = -225\ \mathrm{kJ}$ and $\Delta S = -150\ \mathrm{JK}^{-1}$, convert $\Delta H$ to consistent units: $\Delta H = -225,\!000\ \mathrm{J}$.

Solving $0 = \Delta H - T\Delta S$ yields $T = \frac{\Delta H}{\Delta S} = \frac{-225,\!000\ \mathrm{J}}{-150\ \mathrm{JK}^{-1}} = 1,\!500\ \mathrm{K}$.

The temperature at which $\Delta G = 0$ is therefore $1,\!500\ \mathrm{K}$.

The correct option is A.

Hence, the correct answer is option: 'A'.

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